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Force-Deflection Curve Data

Force-Deflection Curve Data

Formula Expression

Parameters

SymbolNameUnit
DeDemm
DiDimm
h0h0mm
materialmaterial
ttmm

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Detailed Calculation Guide

DIN 6796 Force‑Deflection Curve Data Generation

1. Generation Objective

To provide designers with complete force‑deflection scan data for DIN 6796 disc spring washers within the range 0 ≤ s ≤ h₀, for plotting characteristic curves, stiffness analysis, preload‑compression matching, and interfacing with VDI 2230 system.

Data is based on the classic Almen‑Laszlo formula, applicable to smooth conical surface, toothless DIN 6796 washers.


2. Calculation Formula Review

Dimensionless compression $\delta = s/t$, cone height ratio $\eta = h_0/t$, load formula:

$$F(s) = \frac{4E}{1-\nu^2} \cdot \frac{t^4}{K_1 D_e^2} \cdot \delta \left[ (\eta - \delta)\left(\eta - \frac{\delta}{2}\right) + 1 \right]$$

Where shape coefficient:

$$K_1 = \frac{1}{\pi} \cdot \frac{\left(\dfrac{c-1}{c}\right)^2}{\dfrac{c+1}{c-1} - \dfrac{2}{\ln c}}, \quad c = \frac{D_e}{D_i}$$

3. Method for Generating Scan Data

  1. Obtain washer geometry and material parameters:
    $D_e, D_i, t, h_0, E, \nu$

(typically MPa, ).

  1. Calculate constants:
  2. Outer‑to‑inner diameter ratio $c = D_e/D_i$
  3. $K_1$ using the above formula
  4. Constant factor $C_F = \dfrac{4E}{1-\nu^2} \cdot \dfrac{t^4}{K_1 D_e^2}$

  5. Set compression sequence:
    From $s=0$ to $s = h_0$ (fully flattened), step size can be $0.05h_0$ or finer.
    If only the elastic working range is needed, limit to $s \le 0.75h_0$.

  6. Calculate $F(s)$ point by point:
    For each $s$, compute $\delta = s/t$, substitute into the dimensionless formula to obtain $F$.

  7. Output table ($s$ [mm], $F$ [N]), for plotting or further analysis.


4. Complete Example: DIN 6796 Washer for M10

4.1 Parameter Selection

Parameter Value Unit
Outer diameter $D_e$ 20.0 mm
Inner diameter $D_i$ 10.2 mm
Thickness $t$ 1.5 mm
Free cone height $h_0$ 1.0 mm
Elastic modulus $E$ 206 000 MPa
Poisson's ratio $\nu$ 0.3

4.2 Constant Calculation

  • $c = 20.0 / 10.2 \approx 1.9608$
  • $\ln c \approx 0.6733$
  • $\dfrac{c-1}{c} = \dfrac{0.9608}{1.9608} \approx 0.4900$
  • $\left(\dfrac{c-1}{c}\right)^2 \approx 0.2401$
  • $\dfrac{c+1}{c-1} = \dfrac{2.9608}{0.9608} \approx 3.081$
  • $\dfrac{2}{\ln c} \approx \dfrac{2}{0.6733} \approx 2.971$
  • Denominator = $3.081 - 2.971 = 0.110$
  • $K_1 = \dfrac{1}{\pi} \cdot \dfrac{0.2401}{0.110} \approx 0.6947$
  • $\dfrac{4E}{1-\nu^2} = \dfrac{4\times206000}{1-0.09} = \dfrac{824000}{0.91} \approx 905\,495\ \text{MPa}$
  • $C_F = 905\,495 \times \dfrac{1.5^4}{0.6947 \times 20^2} = 905\,495 \times \dfrac{5.0625}{277.88} \approx 905\,495 \times 0.018215 \approx 16\,497\ \text{N/mm (note unit correction)}$

More precisely retaining intermediate values:
$t^4 = 5.0625\ \text{mm}^4$, $K_1 D_e^2 = 0.6947 \times 400 = 277.88$
$\dfrac{t^4}{K_1 D_e^2} = 0.018215$
$C_F = 905\,495 \times 0.018215 = 16\,497\ \text{N}$ (this is the force when $\delta=1$ and bracket=1? No, it must be multiplied by $\delta$ and the bracket. Actually $C_F$ is the $F$ when $\delta=1$ and bracket=1, but $\delta=1$ is only possible when $t=h_0$. This is just a coefficient.)

For actual force calculation:

$$F = 16\,497 \times \delta \times [(\eta-\delta)(\eta-\delta/2)+1]$$

where $\eta = h_0/t = 1.0/1.5 = 0.6667$.

4.3 Force‑Deflection Scan Data Table

Step size $\Delta s = 0.1\ \text{mm}$, from $s=0$ to $s=1.0\ \text{mm}$.

$s$ (mm) $\delta = s/t$ Bracket term $F$ (N)
0.00 0.0000 1.0000 0
0.10 0.0667 1.0400 1 144
0.20 0.1333 1.0741 2 365
0.30 0.2000 1.1022 3 639
0.40 0.2667 1.1244 4 956
0.50 0.3333 1.1407 6 277
0.60 0.4000 1.1511 7 601
0.70 0.4667 1.1556 8 928
0.80 0.5333 1.1541 10 256
0.90 0.6000 1.1467 11 587
1.00 0.6667 1.1333 12 922

($F$ values in the table are rounded; exact values can be obtained via calculator.)

Note: The flattening force $F_{flat}$ is the force at $s = h_0 = 1.0\ \text{mm}$. In this example, $F_{flat} \approx 12.9\ \text{kN}$. This differs from the earlier quick estimate of 11 kN due to rounding in the $C_F$ calculation. This table uses more precise intermediate values:
- Exact $C_F$ = $905\,495 \times (5.0625 / 277.88) = 16\,497$
- $F_{flat} = 16\,497 \times 0.6667 \times 1.1333 = 16\,497 \times 0.7556 \approx 12\,465\ \text{N}$ (differs from table value 12 922 because the bracket term at $\delta=0.6667$ should actually be $(\eta-\delta)(\eta-\delta/2)+1 = (0)(0.3333)+1 = 1.0$!)
Ah, error: when $\delta=\eta=0.6667$, $\eta-\delta=0$, therefore bracket = 1. So $F_{flat} = 16\,497 \times 0.6667 \times 1 = 10\,998\ \text{N}$. The bracket term in the last row of the table should be 1.0, not 1.1333. The previous bracket calculation mistakenly used $\delta=0.6667$ without zeroing. Correct calculation is as follows:

  • $s=1.0\ \text{mm}, \delta=0.6667$, $\eta-\delta=0$, then bracket = $0 \times (0.6667-0.3333) + 1 = 1$. Therefore $F = 16\,497 \times 0.6667 \times 1 = 10\,998\ \text{N}$.
  • The bracket terms in the preceding rows of the table should also be recalculated.

Below is the corrected accurate scan data, step size 0.1 mm, precise to 1 N.


4.4 Corrected Standard Data Table

$s$ (mm) $\delta$ $(\eta-\delta)(\eta-\delta/2)+1$ $F$ (N)
0.00 0.0000 1.0000 0
0.10 0.0667 $(0.6000)\times(0.6333)+1 = 1.3800$ $16\,497 \times 0.0667 \times 1.3800 = 1\,520$
0.20 0.1333 $(0.5334)\times(0.6000)+1 = 1.3200$ $16\,497 \times 0.1333 \times 1.3200 = 2\,904$
0.30 0.2000 $(0.4667)\times(0.5667)+1 = 1.2644$ $16\,497 \times 0.2000 \times 1.2644 = 4\,171$
0.40 0.2667 $(0.4000)\times(0.5334)+1 = 1.2133$ $16\,497 \times 0.2667 \times 1.2133 = 5\,339$
0.50 0.3333 $(0.3334)\times(0.5000)+1 = 1.1667$ $16\,497 \times 0.3333 \times 1.1667 = 6\,414$
0.60 0.4000 $(0.2667)\times(0.4667)+1 = 1.1244$ $16\,497 \times 0.4000 \times 1.1244 = 7\,422$
0.70 0.4667 $(0.2000)\times(0.4334)+1 = 1.0867$ $16\,497 \times 0.4667 \times 1.0867 = 8\,364$
0.80 0.5333 $(0.1334)\times(0.4000)+1 = 1.0533$ $16\,497 \times 0.5333 \times 1.0533 = 9\,272$
0.90 0.6000 $(0.0667)\times(0.3667)+1 = 1.0244$ $16\,497 \times 0.6000 \times 1.0244 = 10\,144$
1.00 0.6667 $(0.0000)\times(0.3333)+1 = 1.0000$ $16\,497 \times 0.6667 \times 1.0000 = 10\,998$

This data can be directly imported into Excel or MATLAB to plot the $F-s$ curve. Curve characteristics: initial stiffness is low, force increases slowly with compression, stiffness decreases slightly near flattening.


5. Plotting and Design Application

  • Recommended elastic working range: $s = 0.15\sim0.75h_0$ (in this example 0.15~0.75 mm), corresponding force values approximately 2 000~9 000 N, stiffness variation is gentle in this range.
  • Flattening force safety check: Maximum working preload $F_{Mmax} \le F_{flat}/1.3$ (approximately 8 460 N). According to this example, preload should be ≤ 8 460 N. From the table, the corresponding $s \approx 0.72\ \text{mm}$, satisfying ≤ 0.75h₀.
  • Stiffness extraction: Take values near the working point, calculate secant stiffness $k_W \approx \Delta F/\Delta s$, for use in VDI 2230 compliance calculations.

6. Parametric Generation Recommendations

In practical use, a small program can be written, inputting $D_e, D_i, t, h_0$, automatically calculating $K_1$ and $C_F$, then outputting the $s-F$ data table with any desired step size. Note: - When $s>h_0$, the washer is already flattened, the formula no longer applies, and the force will rise sharply (approaching solid compression). - For toothed DIN 6796 variants (non‑standard), a correction factor must be applied.

Conclusion:
Using the Almen‑Laszlo formula, complete force‑deflection scan data for DIN 6796 washers from free to flattened state can be generated, providing accurate elastic characteristic curves for connection design. The example data table provided in this document can be directly used for calibration and selection, avoiding repeated chart lookups.

$E=206\,000$$\nu=0.3$

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